Is there a difference between allowing only countable unions/intersections, and allowing arbitrary (possibly uncountable) unions/intersections? As in the title, I am asking if there is a difference between allowing set-theoretic operations over arbitrarily many sets, and restricting to only countably many sets.
For example, the standard definition of an topology on a set $X$ requires that arbitrary unions of open sets are open.  Do I lose anything significant if I restrict this to just unions of countably many (open) sets?
I cannot come up with an example where it makes a difference.
 A: Yes, it makes a difference. Here’s a very simple example. Let $D$ be any uncountable set; take it to be $\Bbb R$, if you want to be very specific. Let $\mathscr{A}=\{X\}\cup\{A\subseteq D:A\text{ is countable}\}$. The family $\mathscr{A}$ is closed under finite intersections and countable unions, but it’s not a topology, because it’s not closed under arbitrary unions. In particular, if $F$ is any non-empty finite subset of $D$, then $D\setminus F\notin\mathscr{A}$. However, the topology on $D$ generated by $\mathscr{A}$ is the discrete topology, in which every subset of $D$ is open, since every subset of $D$ is the union of countable (indeed finite) subsets of $D$: if $A\subseteq D$, then $$A=\bigcup_{x\in A}\{x\}\;.$$
A: Let $X$ be an uncountable set. Let 
\[ 
  \tau = \{ O \subseteq X \mid O = X \text{ or } O \text{ is at most countable}\}
\]
Then $\tau$ contains $\emptyset$ and $X$, is closed under finite intersections and under countable unions. But it isn't a topology on $X$ as it isn't closed under arbitrary unions. So it makes a difference.
A: *Edit:*I originally let $X$ be just a Hausdorff space, but in that case the two aren't guaranteed to be different.
Here's an example to see a where it makes a difference.  Let $X$ be an uncountable Polish space.  If you look at the smallest collection of subsets of $X$ that a) contains all the open sets of $X$ and b) is closed under complements and countable unions (hence also countable intersections), you get the Borel $\sigma$-algebra of $X$.
But if you look at the smallest collection of subsets of $X$ that a) contains all the open sets and b) is closed under complements and arbitrary unions (hence also arbitrary intersections), this is $\cal{P}$$(X)$. This is because it includes all closed sets, hence all singletons, and then we can take an arbitrary union to get any subset of $X$.
